help-27

[manic sedge]

Same arc is equal which arc you're talking about

[idle swift]

ye thats what im trying to essentially find out

i just gave you a way to think

it might/ might not work

[manic sedge]

I could prove alpha beta B = alpha beta c

[idle swift]

well its 90 degrees no?

[untold lance]

Maybe we can use coordinate geometry here..?

[manic sedge]

I checked the solution now

They just used simple geometry

So confusing

Anybody could explain

[untold lance]

oh I got it

what they did is

They assumed that the bisector of BAC, AE, meets the circum circle of ABC at E

[manic sedge]

Ohk

[untold lance]

now, they dropped a perpendicular from E on BC

and they proved DE is the bisector of BC

then it should be the perpendicular bisector of BC

so

[manic sedge]

So?

[untold lance]

bisector of angle BAC and perpendicular bisector of BC meet at a point of the circumcircle

namely, E

statement is proved

[manic sedge]

Oh

Thanks girl 🫂🫂@untold lance

[untold lance]

actually they did the opposite

they said DE is the bisector

and then proved it's perpendicular

both ways are valid

[untold lance]

you're welcome.

[manic sedge]

🥲

[manic sedge]

,w plot x^2 + {y -3/4(x^2)^(1/3)}^2 < 1×1

[woven radish]

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[untold lance]

[manic sedge]

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[restive river]

I didn't really understand this one

[restive river]

At what point do I use to measure 2011 to 2016 on the graph

[supple knot]

do you understand this part of the question

[restive river]

Not really

[supple knot]

if t = 0, then the year is 2011 because 0 years after 2011 is still 2011

if t = 1, then the year is 2012 because 1 year after 2011 is 2011 + 1 = 2012

can you find the pattern?

[restive river]

so t=5 will be 2016

I'm literally so smarty

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[fresh cypress]

how would you do this without a calculator?

i can convert it to modulus argument form

[restive river]

Binomial theorem maybe

[fresh cypress]

but then i cant simplify it without a calculator

[restive river]

Or use pascal's triangle

[supple knot]

,tex .demoivre

[woven radish]

riemann

[fresh cypress]

ive done that to get 2^7 (cis (7pi /3) )

[fresh cypress]

maybe but the chapter is on demoivres so I think it wants us to use that to solve it

[supple knot]

why can't you simplify and distribut 2^7 here?

[restive river]

I guess so

[supple knot]

do you know what cos(7pi/3) and sin(7pi/3) equal?

[fresh cypress]

it would be equal to pi/3 I think

as in using that angle instead

[restive river]

Breaking the numerator in the multiple of denominator of the angle should help

[fresh cypress]

i think for sin its sqrt3 /2

and then for cos its a half

[restive river]

Negative half?

As it lies in the 3rd quadrant

[fresh cypress]

I believe its just a half

thats what i remember it as

i mean logically it makes sense

as it goes from 1 to 0 in pi/2

[restive river]

I mean the angle is pi +pi/3

[supple knot]

7pi/3 is in the first quadrant

[supple knot]

pi + pi/3 = 4pi/3

[restive river]

Lol

Yes

[fresh cypress]

ahh yes ive simplified it to 64(1+isqrt(3))

[restive river]

My head thinking 3pi/3 instead of 6pi/3

[fresh cypress]

but the answer is 64(i + sqrt(3))

but that doesnt make sense

[supple knot]

why not

do you know what 2^7 is?

[fresh cypress]

because my sqrt(3)/2 is for the i component

[supple knot]

show this in the book

[fresh cypress]

@supple knot it's Q6

[supple knot]

,calc cos(7pi/3)

[woven radish]

Result:

0.5

[fresh cypress]

The book does usually get things wrong

[supple knot]

show the question too

[supple knot]

oh this

[fresh cypress]

I've seen my error

Oops

I thought it was arctan sqrt (3)

[supple knot]

your angle is wrong

[fresh cypress]

When infact it was the reciprocal of that

Yup

I think that gives pi/6

Rather than pi/3

Got the right answer

Thank you for all your help!

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[pulsar dock]

You don't need to calculate the arctangent if you don't want to. -sqrt(3) -i is clearly not on the unit circle. So divide by the length 2 and the point is at the coordinate (-sqrt(3)/2, -1/2) which you can read right off the unit circle.

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[proud nimbus]

is this correct ?

[polar chasm]

In this step you added (1/3)^2 to left side, but 1/3 to right side

(1/3)^2 is 1/9, so you should add 1/9 to right side

[proud nimbus]

yeah thats what i was thinking what went wrong

thanks

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[unborn crypt]

Hello i have a midterm tomorrow and Im struggling to understand how he found the variance in this example could someone explain me? I tried to look at the formula and I find it complicated

[unborn crypt]

Im trying to get that result on Excel but despite trying Var.S and Covariance.S it doesnt work

please ping me if you're willing to help me because I dont have notifications on

[devout snow]

@unborn crypt Has your question been resolved?

[unborn crypt]

let me try changing channels

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[frigid niche]

Evening guys, just curious about this math problem
Example:
An apartment increases in value from 1,250,000 to 1,475,000 in 3 years. Describe how the apartment's value y USD changes after x years with:

a) Linear relationship

b) Exponential relationship

[devout snow]

@frigid niche Has your question been resolved?

[frigid niche]

@dry robin

[devout snow]

@frigid niche Has your question been resolved?

[untold ore]

If this were linear, the house would increase the same amount each year

If it were exponential, the price of the house would increase relative to the price of the house.

So each year, the price would go up by a larger amount

[frigid niche]

Thanks a lot mate

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[quasi wasp]

Hey I need some help understanding where I went wrong with this integration

[quasi wasp]

The answer is supposed to be a fraction

[tame palm]

Why are you finding the area that is shaded red?

[quasi wasp]

Because I found the area of the rectangle that encloses y = 5 and where the line intercepts the curve. Then I subtracted the red areas from the area of that triangle to find the area that’s inside the curve @tame palm

I hope that makes sense

[tame palm]

That's not the way you should be solving this integral.

[quasi wasp]

How would I solve this?

[tame palm]

You should be evaluating the length of the orange line across the interval from [-1,4] which are the points of intersection of the two equations.

[quasi wasp]

Right ok I’m picking up what you’re saying

How would I do that

[tame palm]

In this integral, what is the length of the orange line?

[quasi wasp]

5 - x^2 - 3x +1?

So x^2 - 3x -4

No

[tame palm]

5 - (x^2 - 3x +1)

The parentheses are important here.

[quasi wasp]

-x^2 + 3x + 4?

[tame palm]

Correct.

And what is the interval for which you need to integrate?

[quasi wasp]

That might work a lot better for me, I’ve just never seen that form of integral because I haven’t done it yet. Also the interval would be -1 to 4

[tame palm]

Correct. All you need to do now is integrate.

[quasi wasp]

Thank you very much

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[dusk sand]

rule below. why are they using ln or log? is it because its the same thing?

[cosmic trail]

wdym? They're just illustrating the change of base formula

[dusk sand]

on left

it says log _b M

then they used

ln

how come

[acoustic vault]

ln(x)=log_e(x)

[cosmic trail]

yes, change of base

they didn't have to use ln

[dusk sand]

ln_e(x) = log_e(x) right

[cosmic trail]

yes although you don't write ln_e

[acoustic vault]

well writing ln_e is kind of redundant

you can completely omit using ln if that is your source of confusion

a lot of people consider log_e(x)=log(x)

[dusk sand]

log_e is not log(x)

[acoustic vault]

well i said a lot of people consider it

[dusk sand]

learned this today so ye

[acoustic vault]

its just convention. when you write log(x) you are choosing not to write the base

[dusk sand]

so why are they illustrating changing base

[acoustic vault]

because it is obvious based on context what base you are talkinga bout

idk it can be important to change base?

[dusk sand]

not base i mean the log vs ln

why are they changing it

[acoustic vault]

if you have an equation with multiple bases

it can be useful to change everything to the same base

[dusk sand]

give me an example

an easy one

lol

[supple knot]

just do problems in your book

[acoustic vault]

$log_5(x)=5*log_7(x)$

[woven radish]

Køter

[acoustic vault]

or yea just do examples

[dusk sand]

wth

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[dusk sand]

ty

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[bitter patrol]

How do u solve the proof

[devout snow]

@bitter patrol Has your question been resolved?

[bitter patrol]

<@&286206848099549185>

[vast rain]

you could always just use the AA similarity criterion

or follow the plan for proof

[bitter patrol]

hmm

so for step 2

i put s is congruent to 2

[bitter patrol]

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[rare mantle]

Yes

Must be an error

Sounds right to me

wait

I think I messed up

yeah

it's not 90

I would make a proportion

$\frac{3}{5} = \frac{150}{x}$

[woven radish]

Kihei

[rare mantle]

and solve for x

although 350 is wrong

[devout snow]

@restive river Has your question been resolved?

[drowsy carbon]

3x = 150×5
x= (150×5)/3
x = 50×5
x = 250

Nvm lol it was solved already

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[final hatch]

just got a quick question, lets say there's 2 triangles, and 2 of 3 sides are proportional on each triangle, so would the 3rd one also be automatically proportional which would make both triangles similar

[past scaffold]

No

But if the angle between them is equal

Then yes

[final hatch]

never mind that

here is the problem

[torn vessel]

they're similar if their angles are all the same

[devout snow]

@final hatch Has your question been resolved?

[wooden wraith]

if the angle between them is congruent, yes

SAS

[final hatch]

so i would have to determine which one of the similarity criteria could be used and if none can be used, they're not similar

[final hatch]

bruh ok how do i do this

pls

[torn vessel]

what criteria can you use to check if these two are similar?

[final hatch]

angles

since it has to add to 180

then the remaining angle for the bigger one would be 60

and the one for the small triangle would be 80

[torn vessel]

good, so are they similar?

[final hatch]

yes, but...

there are 3 triangles

[torn vessel]

now for these two triangles, what criteria can you use?

[final hatch]

sides

side length

i see that 18 x 1 2/3 is 30

and also 12 x 1 2/3 is 18

so there are 2 proportional sides

that means that these two are also similar

[torn vessel]

you also need that the angle between those sides is the same

[final hatch]

how can i tell

[torn vessel]

you just told me the missing angle in the left triangle. And it's marked as 60 in the right triangle.

[final hatch]

so we know one angle is the same but i would need at least 2

[torn vessel]

No, you have the sides are proportional. So you have SAS similarity

[final hatch]

sas requires 2 similar sides and 1 similar angle?

[torn vessel]

two proportional sides and the angle between them to be the same in both triangles.

just two proportional sides does not give you similar triangles.

[final hatch]

but i have one more thing, i need to compare the middle one to the right one

[torn vessel]

the middle and rightmost triangles are both similar to the leftmost triangle.

similarity is transitive

[final hatch]

what does transitive mean?

[final hatch]

wait so does that mean that all 3 are similar?

[torn vessel]

if A and B are similar and B and C are similar, then A and C are similar

[final hatch]

so then all 3 are similar

[torn vessel]

yes

[final hatch]

all right, thanks for your help

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[dawn jay]

How many digits are in the decimal representation of 5^40?

[dawn jay]

my idea is to convert this in to 5^(10*4) but I dont know what to do from there

[devout snow]

@dawn jay Has your question been resolved?

[dawn jay]

<@&286206848099549185> pls help

[jolly merlin]

!status

[devout snow]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[jolly merlin]

@dawn jay

[dawn jay]

2

[jolly merlin]

Do you know about logarithms?

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[weary mural]

hi super quick question

are these 2 steps interchangeable?

[devout snow]

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[fossil garden]

yes

[weary mural]

ok thanks

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[sacred ember]

Can someone explain me why is a different answer with the negative -1/2? Why we use different type of procedures? And why in the 2nd one the power of "k" can be simplified? :)

[devout snow]

@sacred ember Has your question been resolved?

[past scaffold]

$1^k=1$ always
But $(-1)^k$ doesn't have a particular value

[woven radish]

Ｓａｓｕｋｅ

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[keen shuttle]

prof wants an answer in the form of V(t) = a sin( bt ) litres

[keen shuttle]

it says that the cycle is 4 seconds

one full period of sine is 2pi radians

so if i divide that by 4 i can have sin(t*pi/2)

im just wondering where i would put in the 0.5 and 2L

would it be 2+0.5sin(t*pi/2)

or could it be argued that its 2+0.25sin(t*pi/2) because the air is moving in AND out

[devout snow]

@keen shuttle Has your question been resolved?

[lapis stag]

I’d go with that, 2L is the minimum volume, so you’re oscillating between 2 and 2.5 L with each breath

[keen shuttle]

this function oscillates between 1.5 and 2.5L

so should i do 2.25 + 0.25sin(t*pi/2)

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[sacred ember]

Thank you, I was getting confused by mathway bc it throws that -1^k = 1

[sacred ember]

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[mild basin]

Where did the 7 come from?

[hollow pollen]

8 * 2^x - 2^x = 7 * 2^x

8 * 2^x - 1 * 2^x = (8 - 1) * 2^x = 7 * 2^x

[mild basin]

Why?

I don’t get it lol

[winter patrol]

they subtracted 2^x from both sides of the equation

[hollow pollen]

8a - 1a = ...

[mild basin]

Oh..

[mild basin]

7a

[hollow pollen]

exactly

now imagine that a = 2^x

[mild basin]

Is substitution needed? Or not really

Maybe just easier to see with substitution sometimes?

[hollow pollen]

you dont need substitution

[mild basin]

OK

[hollow pollen]

its just easier to see like that

[mild basin]

Thanks

[hollow pollen]

np

[mild basin]

Alright so then what’s going on for the final answer? I don’t understand where the 3- is coming from lol..
Could you also write the final answer as x = log_2(8) - log_2(7)?

[acoustic vault]

yes but log_2(8)=3

2^3=8

[mild basin]

Oh of course

Ty

There are some steps being skipped

Now I see what’s going on

Like you would wanna divide both sides by 7 before taking the log ya?

You can’t take log and then divide by 7 after

AFAIK?

[acoustic vault]

you can not divide by 7

[mild basin]

But that’s what he did

[acoustic vault]

yea but you dont ahve to

[mild basin]

Wdym?

[acoustic vault]

log_2(7 * 2^x)=log_2(7)+log_2(2^x)=log_2(7)+x

then just subtract log_2(7)

[acoustic vault]

oh sorry i mean that you can choose not to

[mild basin]

Ohhh

OK

[acoustic vault]

yea my bad that was bad wording

[mild basin]

How did you get the right side? Log_2(7) + x

[acoustic vault]

im only evaluating the LHS there btw

and just using logarithm rules

[mild basin]

Oh

[acoustic vault]

so the end would be $log_2(7)+x=log_2(8)$

[woven radish]

Køter

[acoustic vault]

and you would just subtract log_2(7) from both sides

[mild basin]

Wouldn’t it be x*log_2(7)?

Oh

Nvm

[acoustic vault]

no cause log(a * b)= loga + logb

[mild basin]

Ya

[acoustic vault]

:)

[mild basin]

Ty I get it now

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[wicked zodiac]

so we got 2x+3 /x-1 > 4

[wicked zodiac]

this is my answer

https://cdn.discordapp.com/attachments/952581582093426769/1083398776431059064/IMG_20230309_163904.jpg

but the solution shows

https://cdn.discordapp.com/attachments/952581582093426769/1083398776758227044/IMG_20230309_163912.jpg

are both answers correct ? bc in the second pic he *(-1) changing the >0 to <0

<@&286206848099549185>

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[wicked zodiac]

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[restive river]

determine the area whose area is given by this lime. How can I spell this expression differently?

[main gull]

You don't need to close and reopen a new channel if no one answers to you

,rotate

[devout snow]

@restive river Has your question been resolved?

[left robin]

i think it diverges

[restive river]

I need to find some regularity in the expression and write it in a different way

[left robin]

well, that can be done

i did some coding

if we assume that my code is correct, then it diverges

at least it looks like that

[restive river]

This is a similar example, I have to do it this way but I don’t know how

[left robin]

,rotate

[magic thicket]

Each term is less than 1/sqrt(3n^2), so by a trivial comparison the limit is less than 1/sqrt(3). Similarly it is more than 1/sqrt(4) = 1/2

[left robin]

ok i did an oopsie in my code as well

[magic thicket]

Yeah right

Just plot(x, seq)

[left robin]

yep

[magic thicket]

Visibly the limit is 1/2

[left robin]

i would say it is a bit higher

oh wait

probably is 1/2

[magic thicket]

I've found it's equal to the sum of the 1/2k (1/sqrt(2n-k) - 1/sqrt(2n+k)) btw, but it doesn't telescope

[devout snow]

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[left robin]

ima try even larger n

[magic thicket]

Start at a higher n or change the bounds as well

[devout snow]

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[dusty valley]

If given only 2 of 3 critical points (i.e., high point, mid-point or low-point) on a graph of a standard sinusoidal function, can you tell right away whether it's cosine or sine without plugging in those cooridnates and testing them?

[fathom thicket]

the problem is that, firstly cosines and sines are just translations of each other

secondly

[dusty valley]

Yes, so I'm looking for the answer of "No" so that I don't feel so stupid. It seems like it should be easier but there are more factors at play I think.

[fathom thicket]

the sinusoidal function can neither be even nor odd

an even one is what you would think of a cosine

and an odd one is what you would think of a sine

but if its neither

then you wouldnt really have an answer

[dusty valley]

Every sinusoidal function is either cosine or sine.

[fathom thicket]

as in

in this case, it would be more like a combination

of cosine and sine

[dusty valley]

No, sorry, that's not part of the answer when this is your model for the graph of a sinusoidal function: A cos/sin(Bx + C) + D

You have to use cosine or sine to get a sinusoidal graph.

[fathom thicket]

you do know that, sin(Bx + C) can be split up using the addition formula, is what im saying

[dusty valley]

Yes, I do know that a sinusoidal function does not mandate a vertical or horizontal shift.

So, leave out factors A and D and you still need to use cosine or sine function to generate sinusoidal graph. There is no combo function that I'm aware.

[fathom thicket]

wait so you want to know whether to use A sin(Bx + C) + D or A cos(Bx + C) + D ?

[dusty valley]

Omg, are you being serious?

[fathom thicket]

some form Acos(Bx) + Csin(Bx) + D is what i mean by combination of cosine and sine. which is the same

which is mathmatically the same as what you have here A cos/sin(Bx + C) + D

[dusty valley]

My original question asked about giving you 2 cartesian coorindates that are points at key angles on a circle. Given 2 points on a unit circle can you determine without being told point blank, from the characteristics of the 2 points given whether it is a cosine or sine function?

[fathom thicket]

hmm, why are you given both the x and y?

[dusty valley]

Okay, I'm not here to be the teacher. Thanks for trying though. My answer is no but I'll skip the confirmation today.

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[fathom thicket]

@dusty valley im geniunely curios though, because you said this

[fathom thicket]

which leads me to believe you are given theta and a y or x coordinate. but now you are given both the x and y coordinate?

if you just wanted to know if a function was f(theta)=cos(theta) or f(theta) = sin(theta), wouldnt you only need to know 1 of these critical points

even without plugging it in

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[dusty valley]

https://imgur.com/u1COCsy

[fathom thicket]

you can tell its cosine then

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[lapis current]

hola?

[lapis current]

alguien español?

english?

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[fossil moth]

i feel like im having a massive brain fart but how does it go from the second line to the last line? when i try subbing everything in, it doesnt look like that

[sick fulcrum]

what did you get

[weak cove]

:0

[fossil moth]

lemme take a pic

😭 like i know im doing something thats obviously stupid but i cant pin it down

could anyone point out whats wrong?

[upper schooner]

Factor both numerator and denom

[fossil moth]

ah

i see

thanks!

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[mild basin]

The last line, not clear, is the differential

Wouldn’t ln4 become 0/4?

I thought differential of ln was u’/u

Oh

I think I might see it.. maybe..

Product rule?

[woven radish]

keto11

[mild basin]

What rule is this?

[final laurel]

it's the constant multiple rule and it's derived from the product rule

[mild basin]

Are we not doing product rule here?

I guess we are just adding 0

[final laurel]

it's the product rule but it's the special case for a constant

[mild basin]

That part was skipped?

So we are multiplying (5x-9) with ln4

[arctic field]

calling this the product rule is a bit misleading

[mild basin]

Why?

[arctic field]

this is a fundamental property of derivatives

[final laurel]

yup, it's a little bit misleading

[arctic field]

(af + g)' = af' + g'

[lusty sapphire]

linearity of derivatives

[mild basin]

d/dx [(5x-9)ln(4)]
= (5)ln(4) + (5x-9)(0/4)(ln(e))

None of that is correct?

I thought I’m doing product rule

[arctic field]

it's correct but done very badly

[mild basin]

lol 😂

Then why the

It should be 🤩

[arctic field]

because the way you calculated the derivative deserves a

ln(4) is a constant

[mild basin]

But it’s correct

Even if it’s crap

[arctic field]

you have (5x - 9)ln(4) = 5ln(4)x - 9ln(4)

[mild basin]

Ya the entire right hand side equals 0

[arctic field]

the derivative is 5ln(4)

[mild basin]

Yes, I was gonna add that underneath but pressed send by mistake

[main gull]

It's not efficient to do product/chain rule honestly

With that problem

[arctic field]

not just product rule but chain rule on ln(4)

[mild basin]

I don’t care about efficiency I just wanna get it correct then I can improve speed afterwards

[arctic field]

you are not thinking about the operations you're doing

you're just pattern matching

it's the wrong approach to maths

[main gull]

It's better to learn the actual fundamentals

[mild basin]

I see multiplication so I use product rule

[arctic field]

[mild basin]

Product rule ftw 🙌

All day

[final laurel]

that's not a bad approach for someone just learning the subject

[main gull]

So $\pi x$ means product rule to you?

[woven radish]

dldh06

[uncut crow]

f(x) = sqrt(x)*sqrt(x)

differentiate f

[mild basin]

I am replacing quotient rule with product rule (negative exponents) and loving it

[mild basin]

No because pi is a constant

I just use power rule

[arctic field]

so is ln(4)

[mild basin]

True

[main gull]

And same thing with thiis

(5x - 9)ln(4)
ln(4) is a constant

[mild basin]

But I thought ln deserves (u’/u)

It’s good to know lnx = 1/x

[mild basin]

So ya it’s pointless as it’s 0

But I’m still new

[main gull]

Also it's not only about efficiency either, it's about accuracy too. Because with

d/dx [(5x-9)ln(4)]
= (5)ln(4) + (5x-9)(0/4)(ln(e))
You have a higher chance of making algebraic mistakes compared to using the fundamentals

[mild basin]

I will get faster eventually

[mild basin]

I got the correct answer tho

Just took a bit longer

[final laurel]

I did the same thing as you, by using the product rule first when I was learning, but it's a lot better to just use the general forms

it makes the problems much simpler

[main gull]

(5x^3 - 9x + 5)ln(23423)
Product rule that

[mild basin]

I would rather see the steps even if they are for constants then skip ahead I think that’s what happened here

But ya

It’s basically a constant

Like pi

Even if it’s in a ln function

So I can apply same rule I suppose

Product rule happens for constants being multiplied it’s just longer way to get there

[final laurel]

$\ln(y) = (5x -9)\ln(4)$
$$\frac{d}{dx}[\ln(y)] = \frac{d}{dx}[\ln(4)(5x-9)]$$
$$\frac{1}{y} \cdot \frac{dy}{dx} = \ln(4) \cdot \frac{d}{dx}[5x-9]$$
$$\frac{1}{y} \cdot \frac{dy}{dx} = 5\ln(4)$$

[woven radish]

keto11

[final laurel]

I guess this would be how you would show steps for this

[mild basin]

(5x^3 - 9x + 5)ln(23423)
Product rule that

(15x^2 - 9)(ln(23423)) + (5x^3 - 9x + 5)(0/23423)

(15x^2 - 9)(ln(23423)) + 0

[arctic field]

product rule f(x) = 1*1*1*1*1*1*1*1*x

[mild basin]

1

[arctic field]

that isn't product rule

[mild basin]

All that shit in front is just 1*x which is just x

I simplified first

Power rule for that one

EZ

[final laurel]

[mild basin]

Makes sense now, with anything not having a variable being a constant

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[undone cobalt]

how do i start it 😭

[grand siren]

,rotate

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[undone cobalt]

i solved it by myself, :p

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[jolly swan]

21 is a factor?

[jolly swan]

+- 21/2 exist?

[winter patrol]

if you meant whether +-21 and +-21/2
are candidates for rational roots, they are.
somehow they're also missing the trivial candidates +-1

[jolly swan]

is always included?

[winter patrol]

+-1? yes

[feral agate]

yes

to be clear: these are just candidates for possible rational roots of the polynomial, not the actual roots

the theorem says that these are the ONLY possible rational candidates

[jolly swan]

I am not entirely clear how it is done is to simplify p and q?

[jolly swan]

okay

[winter patrol]

I am not entirely clear how it is done is to simplify p and q?
wdym

[jolly swan]

what is wdym

oh what do you mean

the p/q rule how to do it

[winter patrol]

identify the factors of the constant term
identify the factors of the leading coefficient

[jolly swan]

what are the factors

[winter patrol]

do you know what factors of an integer are?

e.g. would you be able to tell me the factors of 6

[jolly swan]

yes for example why is +-1

[winter patrol]

well 1 is a factor of every integer

1/1 = 1

[jolly swan]

and +-?

[winter patrol]

wdym

[jolly swan]

where does it come from
+-

[winter patrol]

i'd suggest looking up proof of rational root theorem

that'll prob give you the answers you want

[jolly swan]

Okay thx

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[rugged jewel]

Multivariable calc. Finding possible local extreme points

[rugged jewel]

Im stuck here

[mossy schooner]

multiply the top equation by x^2

[rugged jewel]

Why?

[mossy schooner]

to solve for x

[fathom thicket]

You could've added both equations together ngl. To get 3x² + 6xy + 3y² -27 = 0

Divided by 3

And factorised the bracket to be

(x+y)² - 9 = 0

Also minus both equations aswell

And do the same thing

To get

(x - y)² - 1 = 0

So you'd have 4 system of 2 easily solvable equations of

x + y = ±3

x - y = ±1

[rugged jewel]

I see

What my teacher did was,

Multiply by x^2

Then

Replace x^2 by 'k', and find the zero points in that

Like this

[fathom thicket]

Yh that works, either way you'll have 4 sets of solutions

For my way

x = (±3 + ±1)/2

And y = (±3 - ±1)/2

And then you just test all 4 combinations of + and -

So +3 +1

+3 -1

-3 +1

-3 -1

So (2,1) (1,2) (-1,-2), (-2,-1)

[rugged jewel]

Yep, thx

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[dull otter]

what are the zeroes of the function in this graph?

[warm mesa]

Actually there are 4, as the sign of the function doesn't change around x = -5 its a repeated root

So your factors will be x(x-4)(x-5)²

Hence option B

[warm mesa]

Wait let me rephrase rhat

f(x+5) will be the same graph as f(x) but shifted towards the left by 5, hence it's minimum value will occur at the x for which f is minimum minus 5

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[elder abyss]

I got stuck at this one and need help

[hollow pollen]

try induction

[elder abyss]

I tried tabular method in Integration by parts and couldn't get exactly n!

[supple knot]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[elder abyss]

what is the base then?

[restive river]

n = 1

anyway you don't need to use induction

it helps, but you dont need to

imagine you differentiate by parts n times, what do you get?

what is u? what is dv?

@elder abyss

[elder abyss]

u is x^n and dv is e^-x

[restive river]

yes

[elder abyss]

that's were I got stuck

I couldn't see the pattern

I couldn't get n! in other words

[restive river]

show your work

[elder abyss]

Sorry for the quality

best I can imagine is that it's leading me to -inf. XD

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[devout snow]

@elder abyss Has your question been resolved?

[devout snow]

@elder abyss Has your question been resolved?

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[ebon cloak]

Have enquired regarding this question before; but now my query is limited to the very last step; How do we find the value of |(u-v)/2| having already figured out everything other than the underlined?

[devout snow]

@ebon cloak Has your question been resolved?

[supple knot]

what's theta? the angle between u and v?

[ebon cloak]

Yes.

[supple knot]

How do we find the value of |(u-v)/2|
are you asking where does that vector come from?

or are you asking how to calculate sin(pi/3)?

[ebon cloak]

How do I prove that |(u-v)/2| = sin(pi/3)? Squaring it, applying |a-b|^2 and then taking its root seems wrong.

[supple knot]

u-v is perpendicular to u+v

drawing a picture with u, v, u+v, u-v should help

[pulsar dock]

u-v is not perpendicular to u+v?

[supple knot]

they are if they're unit vectors

https://math.stackexchange.com/questions/1146164/uv-is-orthogonal-to-u-v-if-and-only-if-u-v

[pulsar dock]

Oh I see okay

[supple knot]

yea we've been lazy about hatting variables

[ebon cloak]

Took my smooth brain a while to realise that I could just assume them to be i and j for the sake of visualisation; I understand the perpendicularity of u+v and u-v now but am still not sure how that'll help me find |(u-v)/2|

[supple knot]

did you draw 2pi/3 as the angle between u and v?

in your drawing, there should be a triangle with u, v, and (u-v)/2 with an angle of pi/3 somewhere

[ebon cloak]

As I mentioned; I assume u to be Î and v to be ĵ ; the angle between them is pi/2 and not 2pi/3?

[supple knot]

that's only for visualizing the orthogonality of u+v and u-v. you already solved for the angle between them

[ebon cloak]

🗿

right. my bad.

[ebon cloak]

How do I plot (u-v)/2 ?

[devout snow]

@ebon cloak Has your question been resolved?

[supple knot]

https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/vectors/v/linear-algebra-vector-examples

skip to like 18m

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[devout snow]

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[raw nebula]

hi

[raw nebula]

im so confused on this explanation

here is the question

[hybrid snow]

What's confusing about it

It's just using the taylor series formula

[raw nebula]

how does sqrt3=p3?

[hybrid snow]

It's estimated

[raw nebula]

sqrt3=1.72

so it makes 0 sense

[hybrid snow]

It's an estimation

That's why there's the ≈

[raw nebula]

how do u just estimate 1.72 to 3

wouldnt u use 2?

[hybrid snow]

They didn't

They estimated sqrt(3) as 111/64

[raw nebula]

okay but its a multiple choice question

how do I know what to estimate it as

[hybrid snow]

Because 4 is the closest perfect square to 3

[raw nebula]

bro they estimated it to 3 not 4 tho

[hybrid snow]

So you estimate it around 4

Using a Taylor series around 4

And then plug in 3 into that series

[raw nebula]

but its a multiple choice question dude

[hybrid snow]

Yeah I know

[raw nebula]

how am I supposed to know exactly what to approximate it as

[sick fulcrum]

now you know

it's called learning it

[hybrid snow]

Because the question says to use the "second-degree Taylor polynomial for f around x = 4"

You make a second-degree polynomial

You know what they mean by that right

[raw nebula]

no around x=4 means centered at x=4

and if u use 3 then its not related to the sqrt 3 at all

[hybrid snow]

Yeah but you can use the polynomial to estimate output values for input values around x=4

How is it not related

[raw nebula]

yes but sqrt3 is not around 4

1.72 and 3 are completely different

why didnt they use 2?

[hybrid snow]

No youre confused. They made a Taylor series of sqrt(x) centered at 4

[raw nebula]

yes ik that

[hybrid snow]

Why would they use 2? Sqrt(2) is irrational

[raw nebula]

they didnt use sqrt 2

or sqrt3

they just used 3

[hybrid snow]

They used the fact that $\sqrt{x} ≈ P(x) = 2 + \frac 14 (x-4) + \frac{-\frac{1}{32}}{2!}(x-4)^2$

[woven radish]

Umbraleviathan

[hybrid snow]

The polynomial estimates sqrt(x)

So you just plug in 3 into P(x) to estimate sqrt(3)

[raw nebula]

oh

[hybrid snow]

That's what the literally Taylor polynomial's purpose is supposed to do

To estimate a function around a certain value

[raw nebula]

I thought u would plug in sqrt3

[hybrid snow]

No

[raw nebula]

ive been doing it wrong this whole time. fuck

i must have failed my last test

alr tysm

[hybrid snow]

Np

[devout snow]

@raw nebula Has your question been resolved?

[devout snow]

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[proud zodiac]

hey

[proud zodiac]

can someone help with this

[upper schooner]

What have you tried for this one? any thoughts?

[proud zodiac]

i think its C

but its A...

but how

NVM

dis

[upper schooner]

You got why?

[proud zodiac]

i thought it showed negative 4

on the right

tho why am i not

making everything positive

its +(x)

[upper schooner]

Hmmm what do you mean?

[proud zodiac]

why am i not doing

3xy^2+8x+4

since the + is on the paranthesis

[upper schooner]

Well I guess one way of seeing it is that when you "expand" the brackets there, you just multiply everything inside by 1

[proud zodiac]

butwhat about -8x

wouldnt it go Positive?

[upper schooner]

So you get like +1*(-8x) = -8x, if that makes sense?

[proud zodiac]

oh

OK

This one?

f(x)

What does that mean even

[upper schooner]

Basically for any number x, you get "given back" 3^x

put x in, get given 3^x

[proud zodiac]

2x2x2?

[upper schooner]

So like if you put x=0 in, you would be given 3^0 (= 1) back

[upper schooner]

not quite no

What do you think f(2) would look like from what I've said?

[upper schooner]

as in just in this form here?

[proud zodiac]

= 6

f(2)

Plug 3 in it equals us 6

[upper schooner]

As in don't do any simplifications or anything, what are you supposed to work out, if I get given $x$ I should get back $3^x$, so if I get given 2, what should I get back?

[woven radish]

@upper schooner

[proud zodiac]

um

anyway you Can do the problem

maybe i understnad that better lol

i'm tryna practice my SAT and im so dumb

[upper schooner]

We can't do the problem we set for you, but look at the "similar" one I did for you

[proud zodiac]

pls

[upper schooner]

As before, I said that f(0) is 3^0

And that f(x) is 3^x

[proud zodiac]

ok

[upper schooner]

What happens when you replace x with 2? just that?

[proud zodiac]

oh.

[upper schooner]

Do nothing else but that?

[proud zodiac]

f(2)3^2

[upper schooner]

Cool cool, perfect

Do you know what 3^2 is?

[proud zodiac]

9

[upper schooner]

There you go, that's all!

[proud zodiac]

what happens with f(2)

[upper schooner]

f(2) is 9

[proud zodiac]

kk

u wnna help me with 1 more?

[upper schooner]

Sure thing

[proud zodiac]

-8 right

:p

[upper schooner]

Yea

[proud zodiac]

dude these freka me out

h(x)

[upper schooner]

All they're asking you to do here is solve $-4(x-1) + 2 = -2$

[woven radish]

@upper schooner

[upper schooner]

it's not that scary, it's just the way they wrote it

[proud zodiac]

ok

okay

-4x+1+2=-2

-4x=1

[upper schooner]

Careful with the way you expanded!

[proud zodiac]

expl;ain

pls

[upper schooner]

should be -4x+4+2=-2

[proud zodiac]

howis it +4

ok i c

dang

-4x=4 ?

x=- ?

x=-1 /

[upper schooner]

Erm not quite not quite

How did you get there?

[proud zodiac]

-4x+4+2=-2

4+2=6

-4x+6=-2
-6 -6

-4x=-8

o

-4x=-8

x=2 ?

[upper schooner]

,calc -4(2-1) + 2

[woven radish]

Result:

-2

[upper schooner]

Yep

[proud zodiac]

wait is it -2

or 2

[upper schooner]

2

[proud zodiac]

oki

[upper schooner]

When you put in x=2, you should get out -2

[upper schooner]

which you do

[proud zodiac]

okay

[upper schooner]

You good with all of that?

[devout snow]

@proud zodiac Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[proud zodiac]

hey

[rocky iris]

Factories the left side first

You know how to do that?

[proud zodiac]

yea

x^2-12x=-35

[rocky iris]

Nah

You don't want the constant to be on the left side

OK so do you know the quadratic formula?

[proud zodiac]

no.

a2+b2=c2 ?

[rocky iris]

OK that's the Pythagoras Theorem

OK do you know how to fectorise X2 - 12x + 35?

[proud zodiac]

nooo

can u hlep

[rocky iris]

Ok one sec

So you know the cross method?

[hybrid snow]

Use vietta's theorem

[rocky iris]

Tf is that

[hybrid snow]

One of vieta's equations states that the sum of the roots of a polynomial, ax^2 + bx + c, is simply just -b/a

[rocky iris]

Oh

[hybrid snow]

The product of the two roots is c/a

[devout snow]

@proud zodiac Has your question been resolved?

[eager gate]

Answer is 12 right?

@proud zodiac

[winter patrol]

yes

[devout snow]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[normal lava]

How would you determine if it's monotonic and bounded or not?

[hazy quail]

I think it’s false

[normal lava]

Why

[hazy quail]

Because the nth term is double the n-1th term

[normal lava]

yeah

What about this one

[narrow plank]

Try describing the nth term in the sequence in terms of n and taking the limit as n → ∞

Or try thinking about what values these fractions are approaching (if they approach a value at all)

[normal lava]

Could you explain a little bit more

[narrow plank]

I’m not sure what else to give

[normal lava]

Well what's the definition of converge and diverge?

[narrow plank]

That you can look up

[normal lava]

got it

[narrow plank]

Feel free to check back with your findings

I don’t just wanna give you the answer you know?

I’d love to

[normal lava]

yeah

i think it's false

[narrow plank]

Yeah? What’s your reasoning?

[urban hornet]

Diverges

[urban hornet]

It is false

[narrow plank]

Does it?

[normal lava]

Yeah it diverges

[narrow plank]

As you move further into the sequence, do the terms approach some finite value?

[normal lava]

the infinite sequence of the partial sums of the series does not have a finite limit.

[narrow plank]

You’re not necessarily looking at a series just a sequence

You don’t really care about what the terms add to

[normal lava]

right

[narrow plank]

You care about $\lim_{ n \to \infty} a_n$

[woven radish]

Shen

[narrow plank]

Where $a_n$ is your sequence

[woven radish]

Shen

[normal lava]

right

[narrow plank]

Do you know what the 5th term in your sequence is/looks like?

What about your 10th, 100th, 100000th, and so on?

[normal lava]

the next term is 6/7

Let's do a different example...

[urban hornet]

Same with the harmonic series, the terms get smaller but the series diverges

[normal lava]

yeah i was thinking that too about the harmonic

[narrow plank]

The series diverges, but not the sequence

[normal lava]

you talking about this one or the one before

[narrow plank]

The 1/2, 2/3, 3/4, etc

[normal lava]

ohh

[narrow plank]

This one is a geometric series

[normal lava]

ohhh

Got it right

Thank you guys

[narrow plank]

Np! Close if done